Urysohn’s Lemma:
Proof - Step 1
These notes cover parts of sections 33, 34, and 35. Not
covered is complete regularity.
Urysohn’s Lemma gives a method for constructing a
continuous function separating closed sets.
Urysohn’s Lemma If A and B are closed in a normal
space X , there exists a continuous function f : X → [0, 1]
such that f (A) = {0} and f (B ) = {1}
Note: In a metric space, the following function does this.
f (x ) =
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2
d (x , A) − d (x , B )
d (x , A) + d (x , B )
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Proof - Continued
Order the rationals in [0, 1]: {1, 0, r3 , r4 , . . .}. For each
p ∈ Q construct inductively an open set Up in X so that if
p < r , then Up ⊂ Ur .
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Proof, Continued
To get started, let U1 = X \ B, and U0 (using normality and
open
Aclosed ⊂ U1 ) be an open set containing A with U0 ⊂ U1 .
Then proceed inductively.
Step 2 For p ∈ Q, define Up = 0/ if p < 0 and Up = X if
p > 1.
Step 3 Define f (x ) = inf{p|x ∈ Up }
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Step 4 Check that:
x ∈ Ur ⇒ f (x ) ≤ r
x∈
/ Ur ⇒ f (x ) ≥ r
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Proof - Continued
Tietze’s Extension Theorem
Step 5 Show that f is continuous and has the desired
properties using Step 4 and the fact that if
We will use Urysohn’s Lemma to prove:
c < p < f (x0 ) < q < d and U = Uq \ Up
then f (U ) ⊂ [p, q ] ⊂ (c , d )
Note: If A and B are Gδ sets (countable intesections of
open sets), then the function f from Urysohn’s Lemma can
be constructed so f −1 (1) = B and f −1 (0) = A.
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Tietze’s Extension Theorem: (Extending functions.)
If Aclosed ⊂ X normal and f : A → Y is continuous where Y is
either R or an interval in R, then there exists a continuous
function F : X → Y extending f , i.e. F (a) = f (a) for each a
in A.
and:
Urysohn’s Metrization Theorem: If X is regular and
second countable, then X is metrizable. In fact, X is
homeomorphic to a subspace of I ω .
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